Question 1

$\cos{2x}-3\sin{x}+1=0$ теңдеуiн шешiңiз.

Accepted Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Question 2

Теңдеудi шешiңiз: $\sqrt[5]{x\sqrt{x}}-26=\sqrt{x\sqrt[5]{x}}-82$ .

Accepted Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Question 3

Есептеңіз: $\sqrt[5]{x\sqrt{x}}-26=\sqrt{x\sqrt[5]{x}}-82$ .

Accepted Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Question 4

Боялған фигураның ауданын табыңыз.

Accepted Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

Answer

$\cos{2x}-3\sin{x}+1=0$

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